Electronics Fundamentals And Applications_d. Chattopadhyay And P. C. Rakshit 61e53

<0V and VDS=12V>(VGS−Vp ) , i t

i s v e r i f i e d t h a t t h e JFET i s i n t h e s a t u r a t i o n region of the drain c h a r a c t e r i s t i c s ”) 38 end

Scilab code Exa 13.16.7 To calculate the drain current 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 IDSS =10*10^ -3 // IDSS=s a t u r a t i o n d r a i n c u r r e n t i n 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Ampere when VGS( g a t e −to −s o u r c e v o l t a g e ) =0V Vp = -2 //Vp=p i n c h − o f f v o l t a g e VDD =20 //VDD=d r a i n s u p p l y v o l t a g e RL =1*10^3 //RL=l o a d r e s i s t a n c e i n ohms Rs =2*1000 // Rs= r e s i s t a n c e c o n n e c t e d t o s o u r c e t e r m i n a l i n ohms R1 =12*10^6 //R1= r e s i s t a n c e i n t h e v o l t a g e d i v i d e r n e t w o r k i n ohms R2 =8*10^6 //R2= r e s i s t a n c e i n t h e v o l t a g e d i v i d e r n e t w o r k i n ohms VT =( R2 /( R1 + R2 ) ) * VDD //VT=T h e v e n i n v o l t a g e //VGS=VT−(ID ∗ Rs ) //By S h o c k l e y ’ s e q u a t i o n // IDS=IDSS ∗(1 −(VGS/Vp ) ) ˆ2 where IDS=s a t u r a t i o n d r a i n c u r r e n t t o be c a l c u l a t e d f o r g i v e n v a l u e o f VGS // S u b s t i t u t i n g VGS=(VGS−ID ∗ Rs ) we g e t ID=IDS // ( 1 0 ∗ ID ˆ 2 ) −(101∗ ID ) + 2 5 0 = 0 . . . . . . . . . ( 1 ) where ID i s o b t a i n e d i n mA ID1 =(101+ sqrt ((101^2) -(40*250) ) ) /(2*10) format ( ” v ” ,5) 100

21 ID2 =(101 - sqrt ((101^2) -(40*250) ) ) /(2*10) // ID1 , ID2 a r e

the 2 r o o t s o f the above e q u a t i o n ( 1 ) 22 format ( ” v ” ,5) 23 disp ( ”mA” ,ID1 , ” ID1=” ) 24 disp ( ”mA” ,ID2 , ” ID2=” ) 25 // For ID1 26 VGS = VT -( ID1 * Rs ) //VGS=g a t e −to −s o u r c e

voltage c a l c u l a t e d f o r ID1 27 if ( Vp > VGS ) then 28 disp ( ”mA” ,ID1 , ” As Vp>(VGS c a l c u l a t e d u s i n g ID1 ) , t h e v a l u e r e j e c t e d i s ID1=” ) 29 end 30 disp ( ”mA” ,ID2 , ” T h e r e f o r e , t h e d r a i n c u r r e n t i s =” ) 31 ID = ID2 *10^ -3 // c o n v e r t i n g ID2 i n t e r m s o f Amperes 32 VGS = VT -( ID * Rs ) //VGS=g a t e −to −s o u r c e v o l t a g e 33 disp ( ”V” ,VGS , ”VGS=” ) 34 VDS = VDD -( ID *( RL + Rs ) ) //VDS=d r a i n −to −s o u r c e v o l t a g e 35 format ( ” v ” ,2) 36 disp ( ”V” ,VDS , ”VDS=” ) 37 if ( Vp < VGS & VDS >( VGS - Vp ) ) 38 disp ( ” As Vp=(−2)<(VGS= −0.68V) and VDS=7V>(VGS−Vp ) , i t i s c h e c k e d t h a t t h e JFET o p e r a t e s i n t h e s a t u r a t i o n r e g i o n ”) 39 end

Scilab code Exa 13.16.8 To find the saturation drain current and the minimum value of drain source voltage 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 // For a n−c h a n n e l JFET 7 IDSS =10*10^ -3 // IDSS=s a t u r a t i o n d r a i n c u r r e n t i n

101

8 9 10 11 12 13 14

15 16

Ampere when VGS( g a t e −to −s o u r c e v o l t a g e ) =0V Vp =( -4) //Vp=p i n c h − o f f v o l t a g e VGS =( -2.5) //VGS=g a t e −to −s o u r c e v o l t a g e //By S h o c k l e y ’ s e q u a t i o n IDS = IDSS *(1 -( VGS / Vp ) ) ^2 // IDS=s a t u r a t i o n d r a i n c u r r e n t t o be c a l c u l a t e d f o r g i v e n v a l u e o f VGS format ( ” v ” ,5) disp ( ”mA” , IDS /10^ -3 , ” The d r a i n c u r r e n t f o r g i v e n v a l u e o f VGS i s =” ) // c o n v e r t i n g IDS i n t e r m s o f mA VDSmin = VGS - Vp //VDSmin=minimum v a l u e o f d r a i n −to − source voltage f o r the onset of the s a t u r a t i o n region format ( ” v ” ,5) disp ( ”V” , VDSmin , ” The minimum v a l u e o f VDS f o r s a t u r a t i o n i s =” )

Scilab code Exa 13.16.9 To determine gate source voltage and the transconductance 1 2 3 4 5 6 7 8 9 10 11 12 13 14

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s clc clear VDD =20 //VDD=d r a i n s u p p l y v o l t a g e IDS =0.9 // IDS=d r a i n s a t u r a t i o n c u r r e n t i n t e r m s o f mA Vp = -3 //Vp=p i n c h − o f f v o l t a g e IDSS =8 // IDSS=s a t u r a t i o n d r a i n c u r r e n t i n mA when VGS ( g a t e −to −s o u r c e v o l t a g e ) =0V //By S h o c k l e y ’ s e q u a t i o n // IDS=IDSS ∗(1 −(VGS/Vp ) ) ˆ2 VGS = Vp *(1 - sqrt ( IDS / IDSS ) ) //VGS=g a t e −to −s o u r c e voltage disp ( ”V” ,VGS , ” The g a t e −to −s o u r c e v o l t a g e VGS i s =” ) //gm=(dIDS /dVGS) | VDS=c o n s t a n t where gm= 102

transconductance 15 gm = -((2* IDSS ) / Vp ) *(1 -( VGS / Vp ) ) 16 format ( ” v ” ,5) 17 disp ( ”mS” ,gm , ” The v a l u e o f t r a n s c o n d u c t a n c e i s =” )

Scilab code Exa 13.16.10 To find the gate source voltage 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 IDS = -15 // IDS=d r a i n s a t u r a t i o n c u r r e n t i n t e r m s o f mA 7 Vp =5 //Vp=p i n c h − o f f v o l t a g e 8 IDSS = -40 // IDSS=s a t u r a t i o n d r a i n c u r r e n t i n mA when 9 10 11 12

VGS( g a t e −to −s o u r c e v o l t a g e ) =0V //By S h o c k l e y ’ s e q u a t i o n // IDS=IDSS ∗(1 −(VGS/Vp ) ) ˆ2 VGS = Vp *(1 - sqrt ( IDS / IDSS ) ) //VGS=g a t e −to −s o u r c e voltage disp ( ”V” ,VGS , ” The g a t e −to −s o u r c e v o l t a g e VGS i s =” )

Scilab code Exa 13.16.11 To calculate Rs and the channel resistance 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 IDSS =10*10^ -3 // IDSS=s a t u r a t i o n d r a i n c u r r e n t i n

Ampere when VGS( g a t e −to −s o u r c e v o l t a g e ) =0V 7 Vp = -5 //Vp=p i n c h − o f f v o l t a g e 8 VDD =24 //VDD=d r a i n s u p p l y v o l t a g e 103

9 VDS =8 //VDS=d r a i n −to −s o u r c e v o l t a g e 10 ID =4*10^ -3 // ID=d r a i n c u r r e n t i n Ampere 11 R1 =2*10^6 //R1= r e s i s t a n c e i n t h e v o l t a g e

divider

n e t w o r k i n ohms 12 R2 =1*10^6 //R2= r e s i s t a n c e 13 14 15 16 17 18 19 20 21

in the voltage d i v i d e r

n e t w o r k i n ohms VT =( R2 /( R1 + R2 ) ) * VDD //VT=T h e v e n i n v o l t a g e //By S h o c k l e y ’ s e q u a t i o n // ID=IDS=IDSS ∗(1 −(VGS/Vp ) ) ˆ2 VGS = Vp *(1 - sqrt ( ID / IDSS ) ) //VGS=g a t e −to −s o u r c e v o l t a g e //VGS=VT−(ID ∗ Rs ) where Rs= r e s i s t a n c e c o n n e c t e d a t the source terminal Rs =( VT - VGS ) / ID disp ( ” k i l o ohm” , Rs /10^3 , ” The v a l u e o f Rs =” ) // c o n v e r t i n g Rs i n t e r m s o f k i l o −ohm Rch = VDS / ID // Rch=c h a n n e l r e s i s t a n c e a t t h e Q−p o i n t disp ( ” k i l o ohm” , Rch /10^3 , ” The c h a n n e l r e s i s t a n c e a t t h e Q−p o i n t i s =” ) // c o n v e r t i n g Rch i n t e r m s o f k i l o −ohm

Scilab code Exa 13.16.12 To find the saturation drain current 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 ID =5 // ID=s a t u r a t i o n d r a i n c u r r e n t i n t e r m s o f mA i n

an n−c h a n n e l enhancement mode MOSFET 7 VGS =8 //VGS=g a t e −to −s o u r c e v o l t a g e 8 VT =4 //VT=T h r e s h o l d v o l t a g e 9 VGS2 =10 //VGS2=g a t e −to −s o u r c e v o l t a g e

f o r which s a t u r a t i o n d r a i n c u r r e n t i s t o be c a l c u l a t e d 10 // ID=K∗ (VGS−VT) ˆ2 where K=(IDSS / ( Vp ˆ 2 ) ) and Vp=p i n c h − o f f v o l t a g e , IDSS=d r a i n s a t u r a t i o n c u r r e n t f o r 104

VGS=0 V 11 K = ID /(( VGS - VT ) ^2) 12 ID1 = K *( VGS2 - VT ) ^2 // ID1=The s a t u r a t i o n d r a i n c u r r e n t f o r g a t e −s o u r c e v o l t a g e o f 10V i e VGS2 13 disp ( ”mA” ,ID1 , ” The s a t u r a t i o n d r a i n c u r r e n t f o r g a t e −s o u r c e v o l t a g e o f 10V i s =” )

Scilab code Exa 13.16.13 To calculate drain current along with gate source voltage and drain source voltage 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 // For n−c h a n n e l enhancement mode MOSFET o p e r a t i n g i n 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

active region VT =2 //VT=T h r e s h o l d v o l t a g e K =0.5 //K=(IDSS / ( Vp ˆ 2 ) ) i n t e r m s o f mA/Vˆ2 VDD =15 //VDD=d r a i n s u p p l y v o l t a g e RL =1 //RL=l o a d r e s i s t a n c e i n k i l o ohm R1 =200*10^3 //R1= r e s i s t a n c e i n t h e v o l t a g e d i v i d e r n e t w o r k i n t e r m s o f ohms R2 =100*10^3 //R2= r e s i s t a n c e i n t h e v o l t a g e d i v i d e r n e t w o r k i n t e r m s o f ohms VGS =( R2 /( R1 + R2 ) ) * VDD //VGS=g a t e −to −s o u r c e v o l t a g e disp ( ”V” ,VT , ” T h r e s h o l d v o l t a g e i s =” ) disp ( ”V” ,VGS , ” The g a t e −to −s o u r c e v o l t a g e VGS i s =” ) ID = K *( VGS - VT ) ^2 // ID=d r a i n c u r r e n t i n mA disp ( ”mA” ,ID , ” The v a l u e o f d r a i n c u r r e n t ID i s =” ) VDS = VDD -( ID * RL ) //VDS=d r a i n −to −s o u r c e v o l t a g e disp ( ”V” ,VDS , ” The v a l u e o f d r a i n −to −s o u r c e v o l t a g e VDS i s =” ) if ( VDS >( VGS - VT ) ) then disp ( ” As VDS>(VGS−VT) , ( i . e . 10.5 >(5 −2) ) , t h e 105

o p e r a t i o n i s indeed in the a c t i v e r e g i o n ”) 22 end

Scilab code Exa 13.16.14 To calculate K along with drain current and drain source voltage 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 // For n−c h a n n e l MOSFET o p e r a t i n g i n t h e d e p l e t i o n

mode 7 VDD =18 //VDD=d r a i n s u p p l y v o l t a g e 8 VGS =0 //VGS=g a t e −to −s o u r c e v o l t a g e 9 RL =600 //RL=l o a d r e s i s t a n c e i n ohms 10 IDSS =18 // IDSS=d r a i n s a t u r a t i o n c u r r e n t i n mA f o r 11 12 13 14 15 16 17 18 19 20 21 22 23 24

g a t e −to −s o u r c e v o l t a g e (VGS) =0V Vp = -5 //Vp=p i n c h − o f f v o l t a g e // Assuming t h a t t h e o p e r a t i o n i s i n t h e a c t i v e region // ID=IDS=IDSS ∗(1 −(VGS/Vp ) ) ˆ2 // ID=(IDSS /Vp ˆ 2 ) (VGS−Vp ) ˆ2 K = IDSS /( Vp ^2) disp ( ”mA/Vˆ2 ” ,K , ” The v a l u e o f K i s =” ) ID = IDSS // ID=d r a i n c u r r e n t disp ( ”mA” ,ID , ” S i n c e VGS=0 , t h e v a l u e o f ID=IDSS i s =” ) VDS = VDD -( ID *( RL /10^3) ) //VDS=d r a i n −to −s o u r c e v o l t a g e and a l s o c o n v e r t i n g RL i n t e r m s o f k i l o ohm disp ( ”V” ,VDS , ” The v a l u e o f VDS i s =” ) disp ( ”V” ,Vp , ” P i n c h o f f v o l t a g e Vp i s =” ) disp ( ”V” ,VGS , ” Gate t o s o u r c e v o l t a g e VGS i s =” ) if ( VDS >( VGS - Vp ) ) then disp ( ” As VDS>(VGS−Vp ) , ( i . e .7.5 >(0 −( −5) ) ) , t h e MOSFET i s a c t u a l l y i n t h e a c t i v e r e g i o n ” ) 106

25 end

Scilab code Exa 13.16.15 To calculate the voltage gain and the output resistance 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 // r g i v e n i n t e x t b o o k i s t a k e n a s r d a f t e r w a r d s . 7 8 9 10 11 12 13 14 15 16 17 18 19

Hence r=r d rd =100*10^3 // r d=d r a i n r e s i s t a n c e i n ohms gm =3500*10^ -6 //gm=t r a n s c o n d u c t a n c e i n t e r m s o f A/V ( or S) RL =5*10^3 //RL=l o a d r e s i s t a n c e i n ohms u = rd * gm // u= a m p l i f i c a t i o n f a c t o r AV =( u * RL ) /((( u +1) * RL ) + rd ) //AV=v o l t a g e g a i n format ( ” v ” ,6) disp ( AV , ” The v o l t a g e g a i n i s =” ) Ro = rd /( u +1) //Ro=o u t p u t r e s i s t a n c e e x c l u d i n g RL format ( ” v ” ,5) disp ( ”ohm” ,Ro , ” The o u t p u t r e s i s t a n c e e x c l u d i n g RL i s =” ) Ro1 =( rd * RL ) /( rd +(( u +1) * RL ) ) // Ro1=Ro’= o u t p u t r e s i s t a n c e i n c l u d i n g RL format ( ” v ” ,6) disp ( ”ohm” , floor ( Ro1 ) ,” The o u t p u t r e s i s t a n c e i n c l u d i n g RL i s =” ) // f l o o r f u n c t i o n i s u s e d t o round down t h e v a l u e

Scilab code Exa 13.16.16 To find the small signal voltage gain 107

1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 13 F i e l d −E f f e c t T r a n s i s t o r s 4 clc 5 clear 6 // I n a FET u s e d i n a CS a m p l i f i e r 7 IDSS =4 // IDSS=d r a i n s a t u r a t i o n c u r r e n t i n mA f o r g a t e 8 9 10 11 12 13 14 15

−to −s o u r c e v o l t a g e (VGS) =0V Vp = -3 //Vp=p i n c h − o f f v o l t a g e RL =10 //RL=l o a d r e s i s t a n c e i n k i l o ohms VGS = -0.7 //VGS=g a t e −to −s o u r c e v o l t a g e gmo = -(2* IDSS ) / Vp //gmo=t r a n s c o n d u c t a n c e i n A/V o f a JFET when VGS=0V gm = gmo *(1 -( VGS / Vp ) ) //gm=t r a n s c o n d u c t a n c e AV = - gm * RL //AV=t h e s m a l l s i g n a l v o l t a g e g a i n disp ( AV , ” The s m a l l s i g n a l v o l t a g e g a i n i s =” ) // D e c i m a l term i n t h e a n s w e r d i s p l a y e d i n t e x t b o o k i s i n c o r r e c t a s 2 . 0 4 ∗ 1 0 = 2 0 . 4 and n o t 2 0 . 0 4 .

108

Chapter 14 Integrated Circuits and Operational Amplifiers

Scilab code Exa 14.12.1 To determine the output voltage along with input resistance and the input current 1 2 3 4 5 6 7 8 9 10 11 12 13

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R1 =1000 //R1=i n p u t r e s i s t a n c e i n ohms i n t h e inverting amplifier circuit Rf =3*1000 // Rf=f e e d b a c k r e s i s t a n c e i n ohms v1 =2 // v1=i n p u t v o l t a g e i n t h e i n v e r t i n g t e r m i n a l o f an a m p l i f i e r c i r c u i t vo = -( Rf / R1 ) * v1 // vo=o u t p u t v o l t a g e disp ( ”V” ,vo , ” Output v o l t a g e i s =” ) disp ( ” Output v o l t a g e i s n e g a t i v e a s i t i s t h e c i r c u i t of i n v e r t i n g a m p l i f i e r ”) disp ( ” k i l o ohm” , R1 /1000 , ” I n p u t r e s i s t a n c e Rin=R1 i s =” ) i = v1 / R1 // i =i n p u t c u r r e n t 109

14

disp ( ”mA” ,i *1000 , ” The i n p u t c u r r r e n t i s =” )

Scilab code Exa 14.12.2 To calculate the voltage gain 1 2 3 4 5 6 7 8 9

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R1 =2*1000 //R1=i n p u t r e s i s t a n c e i n ohms i n t h e non− inverting amplifier circuit Rf =5*1000 // Rf=f e e d b a c k r e s i s t a n c e i n ohms AV =1+( Rf / R1 ) //AV=v o l t a g e g a i n o f t h e non−i n v e r t i n g amplifier circuit disp ( AV , ” The v o l t a g e g a i n o f t h e g i v e n non−i n v e r t i n g a m p l i f i e r c i r c u i t i s =” )

Scilab code Exa 14.12.3 To calculate the voltage gains for difference and common mode signals along with CMRR 1 2 3 4 5 6 7 8

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear // F i r s t c a s e v1 =40*10^ -6 // v1=v o l t a g e a p p l i e d t o t h e non−i n v e r t i n g input terminal v2 = -40*10^ -6 // v2=v o l t a g e a p p l i e d t o t h e i n v e r t i n g input terminal 110

9 vo =100*10^ -3 // vo=o u t p u t v o l t a g e

f o r the above i n p u t s

v1 and v2 10 // S e c o n d c a s e 11 V1 =40*10^ -6 //V1=v o l t a g e 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

a p p l i e d t o t h e non−i n v e r t i n g input terminal V2 =40*10^ -6 //V2=v o l t a g e a p p l i e d t o t h e i n v e r t i n g input terminal Vo =0.4*10^ -3 //Vo=o u t p u t v o l t a g e f o r t h e a b o v e i n p u t s V1 and V2 disp ( ” I n f i r s t c a s e : ” ) vd = v1 - v2 // vd= d i f f e r e n c e s i g n a l v o l t a g e disp ( ” V ” , vd /10^ -6 , ” vd=” ) vc =( v1 + v2 ) /2 // vc=common mode s i g n a l v o l t a g e format ( ” v ” ,5) disp ( ” V ” , vc /10^ -6 , ” vc=” ) // Output v o l t a g e i s vo =(Ad∗ vd ) +(Ac∗ vc ) where Ad and Ac a r e t h e v o l t a g e g a i n s f o r t h e d i f f e r e n c e s i g n a l and t h e common−mode s i g n a l , r e s p e c t i v e l y Ad = vo / vd //Ad c a l c u l a t e d i n f i r s t c a s e a s common mode s i g n a l vc=0 disp ( Ad , ” V o l t a g e g a i n f o r t h e d i f f e r e n c e s i g n a l i s Ad =” ) disp ( ” I n s e c o n d c a s e : ” ) Vd = V1 - V2 //Vd= d i f f e r e n c e s i g n a l v o l t a g e disp ( ” V ” , Vd /10^ -6 , ”Vd=” ) Vc =( V1 + V2 ) /2 // Vc=common mode s i g n a l v o l t a g e format ( ” v ” ,5) disp ( ” V ” , Vc /10^ -6 , ”Vc=” ) Ac = Vo / Vc // Ac c a l c u l a t e d i n s e c o n d c a s e a s d i f f e r e n c e s i g n a l Vc=0 disp ( Ac , ” V o l t a g e g a i n f o r t h e common−mode s i g n a l i s Ac =” ) CMRR = abs ( Ad / Ac ) //CMRR=Common Mode R e j e c t i o n R a t i o =| Ad/Ac | disp ( CMRR , ”Common Mode R e j e c t i o n R a t i o i s CMRR=” )

111

Scilab code Exa 14.12.4 To find the output voltage of the three input summing amplifier 1 2 3 4 5 6 7 8 9 10 11 12 13

14 15

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R1 =1*1000 //R1=i n p u t r e s i s t a n c e i n ohms i n t h e i n v e r t i n g terminal of the a m p l i f i e r c i r c u i t R2 =200 //R2=i n p u t r e s i s t a n c e i n ohms i n t h e i n v e r t i n g terminal of the a m p l i f i e r c i r c u i t R3 =400 //R3=i n p u t r e s i s t a n c e i n ohms i n t h e i n v e r t i n g terminal of the a m p l i f i e r c i r c u i t Rf =500 // Rf=f e e d b a c k r e s i s t a n c e i n ohms v1 = -5 // v1=i n p u t v o l t a g e i n t h e i n v e r t i n g t e r m i n a l o f an a m p l i f i e r c i r c u i t a t R1 r e s i s t o r v2 =3 // v2=i n p u t v o l t a g e i n t h e i n v e r t i n g t e r m i n a l o f an a m p l i f i e r c i r c u i t a t R2 r e s i s t o r v3 =4 // v3=i n p u t v o l t a g e i n t h e i n v e r t i n g t e r m i n a l o f an a m p l i f i e r c i r c u i t a t R3 r e s i s t o r vo = -((( Rf / R1 ) * v1 ) +(( Rf / R2 ) * v2 ) +(( Rf / R3 ) * v3 ) ) // vo= o u t p u t v o l t a g e f o r i n v e r t i n g summing summing amplifier circuit disp ( ”V” ,vo , ” Output v o l t a g e o f t h e 3− i n p u t summing a m p l i f i e r c i r c u i t i s =” ) disp ( ” Output v o l t a g e i s n e g a t i v e a s i t t h e c i r c u i t o f i n v e r t i n g summing a m p l i f i e r ” )

Scilab code Exa 14.12.5 To find the voltage gain along with the output voltage 112

1 2 3 4 5 6 7 8 9 10 11 12

13 14 15 16 17 18 19 20 21 22 23 24

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R1 =1*1000 //R1=i n p u t r e s i s t a n c e i n ohms i n t h e inverting amplifier circuit Rf =50*1000 // Rf=f e e d b a c k r e s i s t a n c e i n ohms A = -( Rf / R1 ) //AV=v o l t a g e g a i n o f t h e i n v e r t i n g amplifier circuit disp (A , ” The v o l t a g e g a i n o f t h e g i v e n i n v e r t i n g a m p l i f i e r c i r c u i t i s =” ) // v i n =0.5∗ s i n ( 1 0 0 ∗ %pi ∗ t ) // v o u t=A∗ v i n = −50∗0.5∗ s i n ( 1 0 0 ∗ %pi ∗ t ) =−25∗ s i n ( 1 0 0 ∗ %pi ∗ t) disp ( ” I f t h e o p e r a t i o n w e r e e n t i r e l y l i n e a r , t h e o u t p u t v o l t a g e would have b e e n −25∗ s i n ( 1 0 0 ∗ %pi ∗ t ) ”) disp ( ” But s i n c e t h e v o l t a g e s u p p l y i s +−12V , t h e op− amp i s s a t u r a t e d when | v o u t | a t t a i n s 12V” ) // L e t a t t i m e t=to , v o u t=−12V // −12=−25∗ s i n ( 1 0 0 ∗ %pi ∗ t o ) to =(1/(100* %pi ) ) * asin (12/25) format ( ” v ” ,8) disp ( ” s ” ,to , ” t o=” ) disp ( ” Thus o v e r t h e e n t i r e c y c l e , ” ) disp ( ” v o u t =−25∗ s i n ( 1 0 0 ∗ %pi ∗ t ) V when 0<=t<=t o ” ) disp ( ” v o u t=−12V when to<=t <=(0.01 − t o ) ” ) disp ( ” v o u t =−25∗ s i n ( 1 0 0 ∗ %pi ∗ t ) V when ( 0 .0 1 − t o )<=t <=(0.01+ t o ) ” ) disp ( ” v o u t =+12V when ( 0 . 0 1 + t o )<=t <=(0.02 − t o ) ” ) disp ( ” v o u t =−25∗ s i n ( 1 0 0 ∗ %pi ∗ t ) V when ( 0 .0 2 − t o )<=t <=0.02 s e c o n d s ” )

113

Scilab code Exa 14.12.6 To find the output voltage of the differentiator 1 2 3 4 5 6 7 8 9 10 11 12 13

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R =2*1000 //R=f e e d b a c k r e s i s t a n c e i n ohms i n t h e differentiator circuit C =0.01*10^ -6 //C=i n p u t c a p a c i t a n c e i n f a r a d i n t h e differentiator circuit // d v i / d t =1.5V/1ms f o r 0 <1ms ( g i v e n ramp i n p u t signal ) // o u t p u t v o l t a g e o f a d i f f e r e n t i a t o r i s g i v e n a s vo =−RC( d v i / d t ) d =1.5/(10^ -3) // d=d v i / d t =1.5V/1ms vo = - R * C * d disp ( ”V” ,vo , ” Output v o l t a g e o f a d i f f e r e n t i a t o r i s ” ) disp ( ” Hence f o r <1 ms , vo = −0.03V=−30mV. O t h e r w i s e , vo =0V” )

Scilab code Exa 14.12.8 To calculate the output voltage 1 2 3 4 5 6 7 8

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R1 =5*1000 //R1=i n p u t r e s i s t a n c e i n ohms i n t h e g i v e n op−amp c i r c u i t Rf =10*1000 // Rf=f e e d b a c k r e s i s t a n c e i n ohms vi =5 // v i=i n p u t v o l t a g e a t t h e i n v e r t i n g t e r m i n a l o f 114

9 10 11 12

an op−amp V =4 //V=v o l t a g e a t t h e non−i n v e r t i n g t e r m i n a l o f an op−amp //By a p p l y i n g s u p e r p o s i t i o n t h e o r e m Vo =((1+( Rf / R1 ) ) * V ) +( -( Rf / R1 ) * vi ) disp ( ”V” ,Vo , ” Output v o l t a g e Vo o f t h e c i r c u i t i s =” )

Scilab code Exa 14.12.9 To find the differential mode gain 1 2 3 4 5 6 7 8 9 10 11

12 13 14 15 16 17 18 19 20 21 22

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 14 I n t e g r a t e d C i r c u i t s and O p e r a t i o n a l Amplifiers clc clear R1 =10*1000 //R1= r e s i s t a n c e i n ohms R2 =100 //R2= r e s i s t a n c e i n ohms R3 =10*1000 //R3= r e s i s t a n c e i n ohms R4 =10*1000 //R4= r e s i s t a n c e i n ohms R5 =10*1000 //R5= r e s i s t a n c e i n ohms // S i n c e t h e v o l t a g e g a i n s o f t h e OP AMPs a r e i n f i n i t e , t h e v o l t a g e s o f t h e p o i n t s X and Y i n t h e g i v e n f i g u r e a r e V1 and V2 r e s p e c t i v e l y // A p p l y i n g K i r c h h o f f ’ s c u r r e n t law a t X // ( V1/R1 ) +((V1−V) /R3 ) +((V1−V2 ) /R2 ) =0 // A p p l y i n g K i r c h h o f f ’ s c u r r e n t law a t Y // ( ( V2−V) /R4 ) +((V2−V1 ) /R2 ) +((V2−Vo ) /R5 ) =0 // E l i m i n a t i n g V from t h e a b o v e e q u a t i o n s //V2 ∗ ( ( 1 / R2 ) +(1/R4 ) +(1/R5 ) +(R3 / ( R2∗R4 ) ) )−V1 ∗ ( ( 1 / R2 ) +(1/R4 ) +((R3/R4 ) ∗ ( ( 1 / R1 ) +(1/R2 ) ) ) )=Vo/R5 //V2∗R−V1∗ r=Vo/R5 . . . . . . . . . . . . . . . . . . . ( 1 ) R =((1/ R2 ) +(1/ R4 ) +(1/ R5 ) +( R3 /( R2 * R4 ) ) ) r =((1/ R2 ) +(1/ R4 ) +(( R3 / R4 ) *((1/ R1 ) +(1/ R2 ) ) ) ) disp (R , ”R=” ) disp (r , ” r=” ) 115

//R=r from a b o v e c a l c u l a t i o n and i t s a n s w e r displayed 24 // Hence from t h e a b o v e e q u a t i o n ( 1 ) A=Vo / ( V1−V2 ) =−( R5∗R) =−(R5∗ r ) 25 disp ( - R5 *r , ” D i f f e r e n t i a l mode g a i n A=Vo / ( V1−V2 )=” ) 23

116

Chapter 15 Active Filters

Scilab code Exa 15.7.1 To design a Butterworth low filter 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 15 A c t i v e F i l t e r s 4 clc 5 clear 6 fc =1000 // f c=g i v e n cut − o f f f r e q u e n c y i n Hz 7 A = -56 //A=r e q u i r e d g a i n t o be d r o p p e d by t h i s much 8 9 10 11 12 13 14

amount i n dB // Also , A=n o r m a l i z e d g a i n o f B u t t e r w o r t h f i l t e r =|A( jw ) /Ao | f =10*1000 // f=g i v e n f r e q u e n c y i n Hz where t h e n o r m a l i z e d g a i n i s d r o p p e d by g i v e n amount // | A( jw ) /Ao|=( −20) ∗n∗ l o g 1 0 (w/wc ) where n=o r d e r o f the f i l t e r // | A( jw ) /Ao|=( −20) ∗n∗ l o g 1 0 ( f / f c ) n = A /(( -20) * log10 ( f / fc ) ) // n=o r d e r o f B u t t e r w o r t h low− f i l t e r disp ( ceil ( n ) ,” Order o f g i v e n f i l t e r t o be d e s i g n e d i s ( n )=” ) // As n=3 ( from a b o v e c a l c u l a t i o n ) we n e e d c a s c a d i n g o f f i r s t −o r d e r s e c t i o n and s e c o n d −o r d e r s e c t i o n 117

15 // For n=3 16 k =0.5 // k=damping f a c t o r 17 Ao =3 -(2* k ) //Ao=DC g a i n f o r e a c h op−amp i n a g i v e n 18 19 20 21 22 23 24 25

26 27 28 29

B u t t e r w o r t h F i l t e r t o be d e s i g n e d R1 =10*1000 //R1=Assumed r e s i s t a n c e i n ohms //Ao=(R1+R2 ) /R1 R2 =( Ao * R1 ) - R1 // f c =1/(2∗ %pi ∗R∗C) R =1000 //R=Assumed r e s i s t a n c e i n ohms C =1/(2* %pi * R * fc ) format ( ” v ” ,5) disp ( ” The d e s i g n e d v a l u e s o f r e s i s t a n c e and c a p a c i t a n c e f o r a low−p a s s B u t t e r w o r t h f i l t e r a r e : ”) disp ( ” k i l o ohm” , R1 /1000 , ”R1=” ) disp ( ” k i l o ohm” , R2 /1000 , ”R2=” ) disp ( ” k i l o ohm” ,R /1000 , ”R=” ) disp ( ” m i c r o Farad ” ,C /10^ -6 , ”C=” )

Scilab code Exa 15.7.2 To design a first order Butterworth active High filter 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 15 A c t i v e F i l t e r s 4 clc 5 clear 6 Ao =5 //Ao=h i g h f r e q u e n c y g a i n o f a g i v e n 7 8 9 10 11 12

B u t t e r w o r t h a c t i v e HP f i l t e r //Ao=(R1+R2 ) /R1 R1 =1000 //R1=Assumed r e s i s t a n c e i n ohms R2 =( Ao * R1 ) - R1 fc =200 // f c=g i v e n cut − o f f f r e q u e n c y i n Hz // f c =1/(2∗ %pi ∗R∗C) R =5*1000 //R=Assumed r e s i s t a n c e i n ohms 118

f i r s t −o r d e r

13 C =1/(2* %pi * R * fc ) 14 disp ( ” The d e s i g n e d v a l u e s

15 16 17 18

o f r e s i s t a n c e and c a p a c i t a n c e f o r a h i g h −p a s s B u t t e r w o r t h f i l t e r are : ”) disp ( ” k i l o ohm” , R1 /1000 , ”R1=” ) disp ( ” k i l o ohm” , R2 /1000 , ”R2=” ) disp ( ” k i l o ohm” ,R /1000 , ”R=” ) disp ( ” m i c r o Farad ” ,C /10^ -6 , ”C=” )

Scilab code Exa 15.7.3 To design a second order band Butterworth filter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

20

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 15 A c t i v e F i l t e r s clc clear fo =1000 // f o=c e n t r e f r e q u e n c y i n Hz f =100 // f=bandwidth i n Hz //Q=wo/w=Q u a l i t y f a c t o r Q =(2* %pi * fo ) /(2* %pi * f ) C1 =0.02*10^ -6 C2 =0.02*10^ -6 //C1=C2=Assumed C a p a c i t a n c e s i n Farad Ao =2 //Ao=g a i n a t t h e c e n t r e f r e q u e n c y //R1∗C1=Q/ ( wo∗Ao ) f o r a c t i v e band p a s s B u t t e r w o r t h filter wo =2* %pi * fo R1 = Q /( Ao * wo * C1 ) R3 = Q /( wo *(( C1 * C2 ) /( C1 + C2 ) ) ) Rp =1/(( wo ^2) * R3 * C1 * C2 ) R2 =( R1 * Rp ) /( R1 - Rp ) disp ( ” The d e s i g n e d v a l u e s o f r e s i s t a n c e and c a p a c i t a n c e f o r a s e c o n d o r d e r band−p a s s Butterworth f i l t e r are : ”) disp ( ” k i l o ohm” , ceil ( R1 /1000) ,”R1=” ) // f l o o r ( ) and 119

21 22 23 24

c e i l ( ) f u n c t i o n s are used in order to get truncated f l o a t i n g values as per the requirement disp ( ”ohm” , floor ( R2 ) ,”R2=” ) disp ( ” k i l o ohm” , ceil ( R3 /1000) ,”R3=” ) disp ( ” m i c r o Farad ” , C1 /10^ -6 , ”C1=” ) disp ( ” m i c r o Farad ” , C2 /10^ -6 , ”C2=” )

Scilab code Exa 15.7.4 To design a notch filter 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 15 A c t i v e F i l t e r s clc clear fo =400 // f o=c e n t r e f r e q u e n c y i n Hz Q =10 //Q=wo/w=Q u a l i t y f a c t o r C1 =0.1*10^ -6 C2 =0.1*10^ -6 //C1=C2=Assumed C a p a c i t a n c e s i n Farad Ao =2 //Ao=g a i n a t t h e c e n t r e f r e q u e n c y //R1∗C1=Q/ ( wo∗Ao ) f o r a c t i v e band p a s s B u t t e r w o r t h filter wo =2* %pi * fo R1 = Q /( Ao * wo * C1 ) format ( ” v ” ,6) R3 = Q /( wo *(( C1 * C2 ) /( C1 + C2 ) ) ) Rp =1/(( wo ^2) * R3 * C1 * C2 ) R2 =( R1 * Rp ) /( R1 - Rp ) // Assuming a r b i t r a r i l y ( R6/R5 ) =10=a a =10 R6 =10*1000 //R6=Assumed r e s i s t a n c e i n ohms R5 = R6 / a R4 = R5 / Ao disp ( ” The d e s i g n e d v a l u e s o f r e s i s t a n c e and c a p a c i t a n c e f o r a notch f i l t e r are : ”) disp ( ” k i l o ohm” ,( R1 /1000) ,”R1=” ) 120

25 26 27 28 29 30 31

disp ( ”ohm” ,( R2 ) ,”R2=” ) disp ( ” k i l o ohm” ,( R3 /1000) ,”R3=” ) disp ( ”ohm” ,R4 , ”R4=” ) disp ( ” k i l o ohm” ,( R5 /1000) ,”R5=” ) disp ( ” k i l o ohm” ,( R6 /1000) ,”R6=” ) disp ( ” m i c r o Farad ” , C1 /10^ -6 , ”C1=” ) disp ( ” m i c r o Farad ” , C2 /10^ -6 , ”C2=” )

121

Chapter 16 Special Devices

Scilab code Exa 16.10.1 To determine the time period of the sawtooth voltage across capacitor C 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 16 S p e c i a l D e v i c e s 4 clc 5 clear 6 Vp =15 //Vp= f i r i n g v o l t a g e o f a u n i j u n c t i o n 7 8 9 10 11 12 13 14

transistor in Volts VBB =40 //VBB=s o u r c e v o l t a g e i n V o l t s n =( Vp / VBB ) // n= i n t r i n s i c s t a n d − o f f r a t i o disp (n , ” The i n t r i n s i c s t a n d − o f f r a t i o i s ” ) R =50*(10^3) //R= r e s i s t a n c e i n ohms C =2000*(10^ -12) // c=c a p a c i t a n c e i n f a r a d T =( R * C * log (1/(1 - n ) ) ) *(10^6) //T=t i m e p e r i o d o f t h e sawtooth v o l t a g e a c r o s s C format ( ” v ” ,7) disp ( ” m i c r o s e c o n d s ” ,T , ” The t i m e p e r i o d i s ” )

122

Chapter 17 Number Systems Boolean Algebra and Digital Circuits

Scilab code Exa 17.17.1 To determine the binary equivalents 1 2 3 4 5 6 7 8 9 10 11

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear x =25 s = dec2bin ( x ) disp ( ,s , ” 1 B i n a r y e q u i v a l e n t o f 25 i s ” ) y =576 s1 = dec2bin ( y ) disp ( , s1 , ” 2 B i n a r y e q u i v a l e n t o f 576 i s ” )

Scilab code Exa 17.17.2 To determine the decimal equivalent 1

// s c i l a b 5 . 4 . 1 123

2 3 4 5 6 7 8

// Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear s= ’ 1111 ’ x = bin2dec ( s ) disp ( ,x , ” D e c i m a l e q u i v a l e n t o f 1 1 1 1 i s ” )

Scilab code Exa 17.17.3 To convert from binary system to decimal system 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear p =1; // i n i t i a l i s i n g v a r i a b l e s q =1; z =0; b =0; w =0; f =0; format ( ’ v ’ ,18) ; // i n c r e a s i n g t h e p r e c i s i o n t o 18 . bin =11.1101; d = modulo ( bin ,1) ; // s e p a r a t i n g t h e d e c i m a l p a r t and the i n t e g e r part d = d *10^10; a = floor ( bin ) ; // r e m o v i n g t h e d e c i m a l p a r t while (a >0) // l o o p t o t a k e t h e b i n a r y b i t s o f i n t e g e r into a matrix r = modulo (a ,10) ; b (1 , q ) = r ; 124

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43

a = a /10; a = floor ( a ) ; q = q +1; end for m =1: q -1 // m u l t i p l y i n g t h e b i t s o f i n t e g e r w i t h t h e i r p o s i t i o n v a l u e s and a d d i n g c =m -1; f = f + b (1 , m ) *(2^ c ) ; end while (d >0) // l o o p t o t a k e t h e b i n a r y b i t s o f d e c i m a l into a matrix e = modulo (d ,2) w (1 , p ) = e d = d /10; d = floor ( d ) p = p +1; end for n =1: p -1 // m u l t i p l y i n g t h e b i t s o f d e c i m a l w i t h t h e i r p o s i t i o n v a l u e s and a d d i n g z = z + w (1 , n ) *(0.5) ^(11 - n ) ; end z = z *10000; // r o u n d i n g o f t o 4 d e c i m a l v a l u e s z = round ( z ) ; z = z /10000; printf ( ” The d e c i m a l e q u i v a l e n t o f 1 1 . 1 1 0 1 i s = %f ” , f+z)

Scilab code Exa 17.17.4 To convert from decimal system to binary system 1 2 3

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits 125

4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

clc clear q =0; b =0; s =0; format ( ’ v ’ ,18) ; // i n c r e a s i n g t h e p r e c i s i o n t o 18 . a =4.625; d = modulo (a ,1) ; // s e p a r a t i n g t h e d e c i m a l p a r t and t h e integer part a = floor ( a ) ; // r e m o v i n g t h e d e c i m a l p a r t while (a >0) // t a k i n g i n t e g e r p a r t i n t o a m a t r i x and converting into equivalent binary x = modulo (a ,2) ; b = b +(10^ q ) * x ; a = a /2; a = floor ( a ) ; q = q +1; end for i =1:10 // f o r v a l u e s a f t e r d e c i m a l p o i n t converting into binary d = d *2; q = floor ( d ) ; s = s + q /(10^ i ) ; if d >=1 then d =d -1; end end k=b+s; printf ( ” The b i n a r y e q u i v a l e n t o f 4 . 6 2 5 i s = %f ” ,k ) ;

Scilab code Exa 17.17.5 To find the equivalent of given number in a code base 5 1

// s c i l a b 5 . 4 . 1 126

2 3 4 5 6 7 8 9

// Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear dec =263 base =5 s = dec2base ( dec , base ) disp ( ,s , ” E q u i v a l e n t o f 263 i n a c o d e b a s e 5 i s ” )

Scilab code Exa 17.17.6 To perform binary addition corresponding to decimal addition 1 2 3 4 5 6 7 8 9

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear x =2 s=x+x s1 = dec2bin ( s ) disp ( , s1 , ” B i n a r y a d d i t i o n c o r r e s p o n d i n g t o d e c i m a l a d d i t i o n 2+2 i s ” )

Scilab code Exa 17.17.7 To perform binary addition and also to show the corresponding decimal addition // s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits 4 clc 1 2 3

127

5 6 7 8 9 10 11 12 13 14 15 16 17 18

clear x= ’ 11111 ’ y= ’ 1011 ’ z = ’ 101 ’ w = ’ 10 ’ v= ’ 1 ’ s1 = bin2dec ( x ) s2 = bin2dec ( y ) s3 = bin2dec ( z ) s4 = bin2dec ( w ) s5 = bin2dec ( v ) a = s1 + s2 + s3 + s4 + s5 b = dec2bin ( a ) disp ( ,b , ” B i n a r y a d d i t i o n o f 11111+1011+101+10+1 i s ” ) 19 disp ( ,a , ” D e c i m a l e q u i v a l e n t c o r r e s p o n d i n g t o a b o v e binary a d d i t i o n i s ”)

Scilab code Exa 17.17.8 To perform the binary subtraction 1 2 3 4 5 6 7 8 9 10 11 12

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear x= ’ 1101 ’ y = ’ 111 ’ s1 = bin2dec ( x ) s2 = bin2dec ( y ) a = s1 - s2 s = dec2bin ( a ) disp ( ,s , ” B i n a r y s u b t r a c t i o n 1101 −111 i s =” )

128

Scilab code Exa 17.17.9 To obtain the output levels of a silicon transistor for given input levels and to show that circuit has performed NOT operation using positive logic 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

18 19

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits clc clear hFE =30 //hFE=dc c u r r e n t g a i n o f g i v e n s i l i c o n transistor VBE =0.8 //VBE=b a s e −e m i t t e r v o l t a g e d r o p a t s a t u r a t i o n VCE =0.2 //VCE= c o l l e c t o r −e m i t t e r v o l t a g e d r o p a t saturation R1 =15*1000 // r e s i s t a n c e a t t h e b a s e s i d e o f t h e t r a n s i s t o r i n ohms R2 =100*1000 // a n o t h e r r e s i s t a n c e a t t h e b a s e s i d e o f t h e t r a n s i s t o r i n ohms RL =2*1000 // l o a d r e s i s t a n c e a t t h e c o l l e c t o r s i d e o f t h e t r a n s i s t o r i n ohms VCC =10 //VCC= c o l l e c t o r s u p p l y v o l t a g e VBB = -10 //VBB=b a s e s u p p l y v o l t a g e // I f t h e i n p u t l e v e l i s 0 v o l t i e v i =0 , t h e open− c i r c u i t e d base v o l t a g e i s given as VB = VBB *( R1 /( R1 + R2 ) ) disp ( ” For i n p u t l e v e l 0 V : ” ) disp ( ” As a b i a s o f a p p r o x i m a t e l y 0 V i s s u f f i c i e n t to cut o f f a s i l i c o n emitter j u n c t i o n , i t f o l l o w s t h a t t r a n s i s t o r i s c u t o f f when v i =0” ) disp ( ”V” ,VCC , ”When v i =0 , t h e o u t p u t v o l t a g e i s vo=VCC =” ) disp ( ” T h i s i n d i c a t e s t h a t t h e o u t p u t i s i n s t a t e 1 when t h e i n p u t i s i n s t a t e 0 ” ) 129

20 21 22 23 24

25 26 27 28 29

30 31 32 33

//When t h e i n p u t l e v e l i s 10 v o l t i e v i =10 , we have t o show t h a t t h e t r a n s i s t o r i s i n s a t u r a t i o n // The minimum b a s e c u r r e n t f o r s a t u r a t i o n i s g i v e n by iB ( min )=iC /hFE iC =( VCC - VCE ) / RL // c o l l e c t o r c u r r e n t when t h e transistor saturates iB = iC / hFE // iB=iB ( min )=minimum b a s e c u r r e n t f o r s a t u r a t i o n i n mA i1 =(10 - VBE ) / R1 // i 1=c u r r e n t t h r o u g h R1 r e s i s t o r c o n n e c t e d a t t h e b a s e s i d e and h e r e v i =10 i s taken i2 =( VBE - VBB ) / R2 // i 2=c u r r e n t t h r o u g h R2 r e s i s t o r connected at the base s i d e iB1 = i1 - i2 // iB1=a c t u a l b a s e c u r r e n t disp ( ” For i n p u t l e v e l 10 V : ” ) if ( iB1 > iB ) then disp ( ” S i n c e iB>iB ( min ) , i t i s v e r i f i e d t h a t t h e t r a n s i s t o r i s i n s a t u r a t i o n ” ) // iB i n d i c a t e s a c t u a l b a s e c u r r e n t & iB ( min ) i n d i c a t e s minimum b a s e c u r r e n t f o r s a t u r a t i o n disp ( ”V” ,VCE , ”When v i =10 , t h e o u t p u t v o l t a g e i s vo=VCE( s a t )=” ) disp ( ” T h i s i n d i c a t e s t h a t t h e o u t p u t i s i n s t a t e 0 when t h e i n p u t i s i n s t a t e 1 ” ) end disp ( ” O v e r a l l i t h a s b e e n t h u s v e r i f i e d t h a t t h e c i r c u i t h a s p e r f o r m e d t h e NOT o p e r a t i o n ” )

Scilab code Exa 17.17.10 To solve the Boolean expression // s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 17 Number Systems , B o o l e a n A l g e b r a , and Digital Circuits 4 clc 1 2 3

130

5 clear 6 A =0 7 B =0 8 C = bitor (A , B ) // b i t w i s e OR o p e r a t i o n i s p e r f o r m e d 9 disp (C , ” B o o l e a n e x p r e s s i o n C=A+B f o r i n p u t s A=0 and 10 11 12 13 14 15 16 17

B=0 i s ” ) A =1 B =0 C = bitor (A , B ) disp (C , ” B o o l e a n e x p r e s s i o n C=A+B f o r i n p u t s A=1 and B=0 i s ” ) A =1 B =1 C = bitor (A , B ) disp (C , ” B o o l e a n e x p r e s s i o n C=A+B f o r i n p u t s A=1 and B=1 i s ” )

131

Chapter 19 VLSI Technology and Circuits

Scilab code Exa 19.29.1 To find the value of inverse of aspect ratio along with power dissipated and load resistance // s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 19 VLSI T e c h n o l o g y and C i r c u i t s clc clear ID =50*10^ -6 // ID=d r a i n c u r r e n t i n a m p e r e s k =25*10^ -6 // k=ue /D i n A/Vˆ2 VDS =0.25 //VDS=d r a i n −to −s o u r c e v o l t a g e VGS =5 //VGS=g a t e −to −s o u r c e v o l t a g e VTH =1.5 //VTH=t h r e s h o l d v o l t a g e w = ID /( k *( VGS - VTH ) * VDS ) //w=W/L format ( ” v ” ,5) disp (w , ”W/L=” ) P = VDS * ID //P=power d i s s i p a t e d by t h e t r a n s i s t o r disp ( ” m i c r o Watt ” ,P *10^6 , ” The d i s s i p a t e d power i s =” ) VDD =5 //VDD=d r a i n s u p p l y v o l t a g e o f g i v e n NMOS transistor 17 R =( VDD - VDS ) / ID //R=l o a d r e s i s t o r t o be c o n n e c t e d i n s e r i e s with the drain 18 disp ( ” k i l o ohm” ,R /1000 , ” The l o a d r e s i s t a n c e i s =” )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

132

Scilab code Exa 19.29.2 To find the pull up and pull down aspect ratio 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 19 VLSI T e c h n o l o g y and C i r c u i t s clc clear ID =50*10^ -6 // ID=d r a i n c u r r e n t i n a m p e r e s k =25*10^ -6 // k=ue /D i n A/Vˆ2 VDEP =3 l =( k *(( - VDEP ) ^2) ) /(2* ID ) // l =(L/W)=a s p e c t r a t i o o f t h e p u l l −up disp (l , ” P u l l −up ( L/W)=” ) VGS =5 //VGS=g a t e −to −s o u r c e v o l t a g e VTH =1 //VTH=t h r e s h o l d v o l t a g e VDs =4.75 //VDs=t h e d r a i n s o u r c e v o l t a g e o f t h e d e p l e t i o n mode p u l l −up i n s a t u r a t i o n VDD =5 //VDD=d r a i n s u p p l y v o l t a g e o f g i v e n NMOS inverter //L/W=(k ∗ (VGS−VTH) ∗VDS) / ID where L/W=p u l l down aspect ratio l1 =( k *( VGS - VTH ) *( VDD - VDs ) ) / ID // l 1=L/W disp ( l1 , ” P u l l −down ( L/W)=” )

Scilab code Exa 19.29.3 To find the value of inverse of aspect ratio of the PMOS transistor for a symmetrical inverter 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 19 VLSI T e c h n o l o g y and C i r c u i t s 4 clc

133

5 clear 6 w =10 //w=W/L v a l u e o f t h e NMOS t r a n s i s t o r 7 8 9 10 11 12 13 14

i n a CMOS inverter un =1350 // un= e l e c t r o n m o b i l i t y f o r NMOS t r a n s i s t o r i n cmˆ2/V s up =540 // up= e l e c t r o n m o b i l i t y f o r PMOS t r a n s i s t o r i n cmˆ2/V s // (Wpu/ Lpu ) ∗ up ∗ ( VINV−VDD−VTHP) ˆ2=(Wpd/ Lpd ) ∗ un ∗ ( VINV− VTHN) ˆ2 // For a s y m m e t r i c a l i n v e r t e r VINV=(VDD/ 2 ) and VTHN =(−VTHP) // A l s o f o r i n p u t v o l t a g e=VDD/2 b o t h t r a n s i s t o r s operate in saturation region // T h e r e f o r e , up ∗ (Wpu/ Lpu )=un ∗ (Wpd/ Lpd ) w1 =( un * w ) / up //w1=Wpu/ Lpu=W/L v a l u e o f t h e PMOS f o r a symmetrical i n v e r t e r disp ( w1 , ”W/L v a l u e o f t h e PMOS t r a n s i s t o r i n a CMOS i n v e r t e r i s =” )

Scilab code Exa 19.29.4 To determine the maximum permissible number of fan outs // s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 19 VLSI T e c h n o l o g y and C i r c u i t s clc clear f =2*10^9 // f=c l o c k f r e q u e n c y i n Hz VDD =3 //VDD=d r a i n s u p p l y v o l t a g e Cl =1*10^ -12 //C1=l o a d c a p a c i t a n c e i n Farad P =50*10^ -3 //P=maximum power d i s s i p a t i o n c a p a b i l i t y i n W/ s t a g e 10 N = P /( f * Cl * VDD ^2) //N=maximum p e r m i s s i b l e number o f fan outs 11 format ( ” v ” ,5) 1 2 3 4 5 6 7 8 9

134

12 13

disp (N , ”N=” ) disp ( floor ( N ) ,” The maximum p e r m i s s i b l e number o f f a n −o u t s i s ( i n t e g e r j u s t b e l o w a c t u a l v a l u e )=” )

Scilab code Exa 19.29.5 To calculate the channel transit time 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 19 VLSI T e c h n o l o g y and C i r c u i t s 4 clc 5 clear 6 L =3*10^ -6 //L=l e n g t h o f an NMOS p a s s t r a n s i s t o r 7 8 9 10 11

in metres VDS =0.5 //VDS=d r a i n −s o u r c e v o l t a g e u =1400*10^ -4 // u= e l e c t r o n m o b i l i t y i n mˆ2/V s t = L ^2/( VDS * u ) // t=c h a n n e l t r a n s i t t i m e format ( ” v ” ,5) disp ( ” n a n o s e c o n d s ” ,t /10^ -9 , ” The t r a n s i t t i m e i s =” )

Scilab code Exa 19.29.6 To calculate the required metal line width 1 2 3 4 5 6 7 8 9 10 11 12

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 19 VLSI T e c h n o l o g y and C i r c u i t s clc clear y =2 // y=l e n g t h u n i t i n m i c r o m e t r e s W =3* y //W=mimimum m e t a l l i n e w i d t h i n m i c r o m e t r e s disp ( ” m i c r o m e t r e s ” ,W , ”W=” ) n =80 // n=number o f d r i v e n i n v e r t e r s i =0.07 // i =a v e r a g e c u r r e n t r a t i n g s i n m i l l i a m p e r e s I = n * i // I= t o t a l c u r r r e n t drawn by n i n v e r t e r s disp ( ”mA” ,I , ” I=” ) 135

13

14 15 16 17

18 19 20 21

// 1mA p e r m i c r o m e t r e o f aluminium l i n e w i d t h i s t h e maximum s a f e a v e r a g e c u r r e n t an aluminium w i r e can c a r r y . disp ( ” T h i s n e e d s a l i n e a t l e a s t w i d t h o f ” ) disp ( ” m i c r o m e t r e s ” ,I ) if (W > I ) then disp ( ” Above c a l c u l a t e d minimum m e t a l l i n e −w i d t h (W) i s t h u s t h e s a f e w i d t h o f t h e m e t a l l i n e d r i v i n g 80 i n v e r t e r s . ” ) end f =5 // f=number o f f a n o u t l i n e s w = f * W //w=r e q u i r e d m e t a l l i n e w i d t h disp ( ” m i c r o m e t r e s ” ,w , ” The m e t a l l i n e −w i d t h r e q u i r e d t o s u p p l y a f a n −o u t o f 5 l i n e s i s =” )

136

Chapter 20 Cathode Ray Oscilloscope

Scilab code Exa 20.9.1 To determine the transit time along with transverse acceleration and spot deflection 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 20 Cathode Ray O s c i l l o s c o p e clc clear e =1.6*10^ -19 // e=c h a r g e o f an e l e c t r o n Va =1000 //Va= p o t e n t i a l d i f f e r e n c e i n v o l t s m =9.11*10^ -31 //m=mass o f an e l e c t r o n v = sqrt ((2* e * Va ) / m ) // v= a x i a l v e l o c i t y o f an e l e c t r o n l =2*10^ -2 // l = a x i a l l e n g t h o f d e f l e c t i n g p l a t e s i n metre t = l / v // t= t r a n s i t t i m e o f t h e beam t h r o u g h t h e deflecting plates format ( ” v ” ,9) disp ( ” s ” ,t , ” The t r a n s i t t i m e i s =” ) Vd =20 //Vd= p o t e n t i a l d i f f e r e n c e a p p l i e d b e t w e e n t h e d e f l e c t i n g plates in volts s =5*10^ -3 // s=s e p a r a t i o n b e t w e e n t h e p l a t e s i n m e t r e ta =( e * Vd ) /( s * m ) // t a=t h e t r a v e r s e a c c e l e r a t i o n i m p a r t e d t o t h e e l e c t r o n s by t h e d e f l e c t i n g 137

17 18 19 20 21 22 23 24 25

voltage format ( ” v ” ,10) disp ( ”m/ s ˆ2 ” ,ta , ” T r a v e r s e a c c e l e r a t i o n i s =” ) L =25*10^ -2 //L=d i s t a n c e o f t h e CRT s c r e e n from t h e c e n t r e o f the d e f l e c t i n g p l a t e s i n metre d =( l * L * Vd ) /(2* s * Va ) // d= d e f l e c t i o n o f t h e s p o t on t h e CRT s c r e e n format ( ” v ” ,13) disp ( ”cm” ,d *100 , ” S p o t d e f l e c t i o n i s =” ) // d i s c o n v e r t e d i n t e r m s o f cm S = d / Vd // S= d e f l e c t i o n s e n s i t i v i t y format ( ” v ” ,10) disp ( ”mm/V” ,S /10^ -3 , ” D e f l e c t i o n s e n s i t i v i t y i s =” ) // S i s c o n v e r t e d i n t e r m s o f mm/V

Scilab code Exa 20.9.2 To calculate the highest frequency of the deflecting voltage 1 2 3 4 5 6 7 8 9 10 11 12 13 14

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 20 Cathode Ray O s c i l l o s c o p e clc clear e =1.6*10^ -19 // e=c h a r g e o f an e l e c t r o n Va =1000 //Va= p o t e n t i a l d i f f e r e n c e i n v o l t s m =9.11*10^ -31 //m=mass o f an e l e c t r o n v = sqrt ((2* e * Va ) / m ) // v= a x i a l v e l o c i t y o f an e l e c t r o n l =1.5*10^ -2 // l = a x i a l l e n g t h o f d e f l e c t i n g p l a t e s i n metre t = l / v // t= t r a n s i t t i m e o f t h e beam t h r o u g h t h e deflecting plates //T=t i m e p e r i o d o f t h e s i n u s o i d a l d e f l e c t i n g v o l t a g e // tmax=maximum t r a n s i t t i m e // ( 0 . 1 / 3 6 0 ) ∗T=tmax , s i n c e 1 c y c l e c o r r e s p o n d s t o 360 degrees 138

15 T =( t *360) /0.1 16 f =1/ T // f=h i g h e s t f r e q u e n c y o f t h e d e f l e c t i n g v o l t a g e 17 format ( ” v ” ,5) 18 disp ( ” kHz ” ,f /1000 , ” The h i g h e s t f r e q u e n c y o f t h e

d e f l e c t i n g v o l t a g e i s =” )

Scilab code Exa 20.9.3 To find the deflection of the spot and the magnetic deflection sensitivity 1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 20 Cathode Ray O s c i l l o s c o p e clc clear V =1000 //V= p o t e n t i a l d i f f e r e n c e i n v o l t s //B=150 g a u s s ( g i v e n ) B =1.5*10^ -2 //B=m a g n e t i c f i e l d i n t e s l a l =1*10^ -2 // l = a x i a l l e n g t h o f d e f l e c t i n g p l a t e s i n metre L1 =20*10^ -2 // L1=(L+( l / 2 ) )=d i s t a n c e o f t h e f l u o r e s c e n t s c r e e n from t h e c e n t r e o f t h e d e f l e c t i o n system i n metre e =1.6*10^ -19 // e=c h a r g e o f an e l e c t r o n m =9.11*10^ -31 //m=mass o f an e l e c t r o n d = B * sqrt ( e /(2* V * m ) ) * l * L1 // d= d e f l e c t i o n o f t h e s p o t format ( ” v ” ,5) disp ( ”cm” ,d *100 , ” The d e f l e c t i o n o f t h e s p o t i s =” ) Sm = d / B //Sm=m a g n e t i c d e f l e c t i o n s e n s i t i v i t y format ( ” v ” ,5) disp ( ”mm/ g a u s s ” , Sm /10 , ” The m a g n e t i c d e f l e c t i o n s e n s i t i v i t y i s =” )

Scilab code Exa 20.9.4 To calculate the frequency of the signal 139

1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 20 Cathode Ray O s c i l l o s c o p e 4 clc 5 clear 6 sw =10 // sw=sweep w i d t h i n cm 7 n =5/2 // n=number o f c y c l e s g i v e n by v e r t i c a l 8 9 10 11 12 13

deflection plates c = sw / n // c=c e n t i m e t r e s o c c u p i e d by one c y c l e o f signal ct =0.1 // c t= c a l i b r a t e d t i m e b a s e o f CRO i n ms/cm t = ct * c // t=t i m e i n t e r v a l c o r r e s p o n d i n g t o c e n t i m e t r e s o c c u p i e d by one c y c l e o f s i g n a l T = t /5 //T=t i m e p e r i o d o f t h e s i g n a l , s i n c e t h e s c a l e i s 5 times magnified f =1/ T // f=f r e q u e n c y o f t h e s i g n a l disp ( ” kHz ” ,f , ” The f r e q u e n c y o f t h e s i g n a l i s =” )

Scilab code Exa 20.9.5 To find the frequency of the vertical signal 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 20 Cathode Ray O s c i l l o s c o p e clc clear // L e t f v=f r e q u e n c y o f t h e v e r t i c a l s i g n a l s i n kHz // f h=f r e q u e n c y o f t h e h o r i z o n t a l s i g n a l s // Number o f h o r i z o n t a l t a n g e n c i e s=nh // Number o f v e r t i c a l t a n g e n c i e s=nv // f v / f h=nh / nv fh =1 nh =3 nv =4 fv =( nh / nv ) * fh disp ( ”Hz” , fv *1000 , ” The f r e q u e n c y o f t h e v e r t i c a l 140

s i g n a l i s =” )

Scilab code Exa 20.9.6 To find the phase difference between the voltages 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 20 Cathode Ray O s c i l l o s c o p e 4 clc 5 clear 6 // s l o p e o f t h e m a j o r a x i s i s n e g a t i v e ( g i v e n ) 7 A =2.6 //A=The maximum y−d i s p l a c e m e n t 8 vyo =1.1 // vyo=t h e v e r t i c a l d i s p l a c e m e n t 9 sino =( vyo / A ) // o=p h a s e d i f f e r e n c e b e t w e e n t h e two

voltages 10 x = asind ( sino ) 11 disp ( ” As t h e m a j o r a x i s o f t h e e l l i p s e h a s a n e g a t i v e s l o p e , p h a s e d i f f e r e n c e b e t w e e n t h e two v o l t a g e s must l i e b e t w e e n 90 d e g r e e and 180 degree ”) 12 disp ( ” d e g r e e ” ,180 -x , ” T h e r e f o r e , p h a s e d i f f e r e n c e b e t w e e n t h e v o l t a g e s i s =” )

141

Chapter 21 Communication Systems

Scilab code Exa 21.13.1 To calculate the critical frequencies and the maximum frequencies 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 21 Communication S y s t e m s 4 clc 5 clear 6 Npe =6*10^10 // Npe=peak e l e c t r o n c o n c e n t r a t i o n 7 8 9 10 11 12 13 14

E l a y e r i n mˆ−3 Npf =10^12 // Npf=peak e l e c t r o n l a y e r i n mˆ−3 fCE =9* sqrt ( Npe ) // fCE= c r i t i c a l layer format ( ” v ” ,5) disp ( ”MHz” , fCE /10^6 , ” C r i t i c a l l a y e r i s =” ) fCF =9* sqrt ( Npf ) // fCF= c r i t i c a l layer format ( ” v ” ,5) disp ( ”MHz” , fCF /10^6 , ” C r i t i c a l l a y e r i s =” ) R =6400 //R=r a d i u s o f t h e e a r t h 142

f o r the

concentration f o r the F frequency f o r the E

frequency f o r the E frequency f o r the F

frequency f o r the F i n km

15 He =110 // He=h e i g h t

o f the E l a y e r above the e a r t h

s u r f a c e i n km 16 ime = asind ( R /( R + He ) ) // ime=a n g l e c o r r e s p o n d i n g t o

maximum f r e q u e n c y fmE f o r E l a y e r i n d e g r e e s 17 format ( ” v ” ,3) 18 fmE = fCE * secd ( ime ) // fmE=maximum f r e q u e n c y 19 20 21 22 23 24

reflected

from t h e E l a y e r disp ( ”MHz” , fmE /10^6 , ” The maximum f r e q u e n c y r e f l e c t e d from t h e E l a y e r i s =” ) Hf =250 // Hf=h e i g h t o f t h e F l a y e r a b o v e t h e e a r t h s u r f a c e i n km imf = asind ( R /( R + Hf ) ) // i m f=a n g l e c o r r e s p o n d i n g t o maximum f r e q u e n c y fmF f o r F l a y e r i n d e g r e e s format ( ” v ” ,3) fmF = fCF * secd ( imf ) // fmF=maximum f r e q u e n c y r e f l e c t e d from t h e F l a y e r disp ( ”MHz” , fmF /10^6 , ” The maximum f r e q u e n c y r e f l e c t e d from t h e F l a y e r i s =” )

Scilab code Exa 21.13.2 To find the maximum distance between the transmitting and receiving points 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 21 Communication S y s t e m s 4 clc 5 clear 6 R =6400 //R=r a d i u s o f t h e e a r t h i n km 7 He =110 // He=h e i g h t o f t h e E l a y e r a b o v e t h e e a r t h

s u r f a c e i n km 8 ime = asin ( R /( R + He ) ) // ime=a n g l e c o r r e s p o n d i n g t o maximum f r e q u e n c y fmE f o r E l a y e r i n r a d i a n 9 format ( ” v ” ,10) 10 o =( %pi /2) - ime // o=a n g l e made by t h e i n c i d e n t r a y a t the centre of the earth in degrees 143

11 format ( ” v ” ,5) 12 L =2* o * R //L=maximum d i s t a n c e b e t w e e n t h e t r a n s m i t t i n g

and t h e r e c e i v i n g p o i n t s on t h e e a r t h s u r f a c e f o r s i n g l e hop t r a n s m i s s i o n o f t h e r a d i o w a v e r e f l e c t e d from t h e E l a y e r 13 format ( ” v ” ,5) 14 disp ( ”km” ,L , ” The maximum d i s t a n c e f o r s i n g l e hop t r a n s m i s s i o n i s =” ) 15 // Answer g i v e n i n t e x t b o o k i s 2 4 5 9 km which i s i n c o r r e c t a s i t i s a c t u a l l y a r o u n d 2 3 5 6 km .

Scilab code Exa 21.13.3 To find the height of the point above the ground from which the wave is reflected back 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 21 Communication S y s t e m s 4 clc 5 clear 6 // f c =9∗ s q r t ( Np ) 7 fc =3*10^6 // f c= c r i t i c a l f r e q u e n c y i n Hz 8 Np =( fc ^2) /81 //Np= e l e c t r o n c o n c e n t r a t i o n a t t h e 9 10 11 12 13

r e f l e c t i n g point // h=h e i g h t o f t h e r e f l e c t i n g p o i n t from t h e bottom of the l a y e r //Np= ( 5 ∗ 1 0 ˆ 1 0 ) +(10ˆ9∗ h ) . . . . ( g i v e n ) h =( Np -(5*10^10) ) /10^9 H =100 //H=h e i g h t a b o v e t h e s u r f a c e o f t h e e a r t h i n km disp ( ”km” ,h +H , ” The r e q u i r e d h e i g h t a b o v e t h e g r o u n d i s =” )

144

Chapter 23 Lasers Fibre Optics and Holography

Scilab code Exa 23.31.1 To calculate the number of photons emitted per second 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y clc clear y =630*10^( -9) // y=e m i t t e d w a v e l e n g t h i n m e t e r s c =3*10^8 // c= v e l o c i t y o f l i g h t i n f r e e s p a c e i n m/ s v = c / y // v=f r e q u e n c y o f t h e e m i t t e d r a d i a t i o n format ( ” v ” ,9) disp ( ” The f r e q u e n c y o f t h e e m i t t e d r a d i a t i o n i s ” ) disp ( ”Hz” ,v , ” v=” ) h =6.62*10^( -34) // h=Planck ’ s c o n s t a n t P =1*10^( -3) //P=o u t p u t power o f g a s l a s e r ( g i v e n ) n = P /( h * v ) format ( ” v ” ,9) disp ( ” s ˆ−1” ,n , ” The number o f p h o t o n s e m i t t e d p e r s e c o n d i s =” )

145

Scilab code Exa 23.31.2 To calculate the coherence time and the longitudinal coherence length 1 2 3 4 5 6 7 8 9 10 11

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y clc clear V =500 //V=bandwidth o f a He−Ne l a s e r i n Hz t =1/ V // t=c o h e r e n c e t i m e disp ( ”ms” ,( t *(10^3) ) ,” The c o h e r e n c e t i m e i s =” ) c =3*10^8 // c= v e l o c i t y o f l i g h t i n m/ s Lc = c / V // Lc= l o n g i t u d i n a l c o h e r e n c e l e n g t h disp ( ”km” ,( Lc /1000) ,” The l o n g i t u d i n a l c o h e r e n c e l e n g t h i s =” )

Scilab code Exa 23.31.3 To calculate the minimum difference between two arms of a Michelson interferometer 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y 4 clc 5 clear 6 //To o b t a i n i n t e r f e r e n c e f r i n g e s o f good v i s i b i l i t y

t h e p a t h d i f f e r e n c e f o r t h e c e n t r a l f r i n g e must be an i n t e g r a l m u l t i p l e o f e a c h o f t h e 2 wavelengths . 7 // 2∗ d=(n1 ∗ y1 ) =(n2 ∗ y2 ) where y1 & y2 a r e 2 wave− l e n g t h s and d=p a t h d i f f e r e n c e and n1 and n2 a r e 2 integers

146

8 9 10 11 12

// ( 2 ∗ d ) ∗ ( ( 1 / y2 ) −(1/ y1 ) ) =(n2−n1 )=m where m i s a n o t h e r integer //Now m=(−2∗d∗Y) / ( y ˆ 2 ) =(2∗ d∗V) / ( v ∗ y ) =(2∗ d∗V) / c =(2∗ d ) / Lc Lc =600 // Lc=c o h e r e n c e l e n g t h i n km d =( Lc /2) // d=minimum d i f f e r e n c e b e t w e e n t h e 2 arms o f the Michelson i n t e r f e r o m e t e r disp ( ”km” ,d , ” The minimum d i f f e r e n c e b e t w e e n t h e two arms o f t h e M i c h e l s o n i n t e r f e r o m e t e r i s =” )

Scilab code Exa 23.31.4 To show that emission for a normal optical source is predominantly due to spontaneous transitions 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y 4 clc 5 clear 6 h =6.62*10^( -34) // h=Planck ’ s c o n s t a n t 7 v =3*10^8 // v= v e l o c i t y o f l i g h t ( a s n o r m a l o p t i c a l

s o u r c e i s m e n t i o n e d ) i n m/ s kB =1.38*10^ -23 //kB=Boltzmann ’ s c o n s t a n t T =1000 //T=t e m p e r a t u r e i n K e l v i n w =6000 //w=w a v e l e n g t h i n Armstrong R =( exp (( h * v ) /( w *(10^ -10) * kB * T ) ) ) -1 //R=t h e r a t i o o f t h e number o f s p o n t a n e o u s t o s t i m u l a t e d transitions 12 disp ( ,R , ”R=” ) 13 if (R >1) then

8 9 10 11

14 15

disp ( ” As t h e r a t i o o f t h e number o f s p o n t a n e o u s t o s t i m u l a t e d t r a n s i t i o n s (R) i s >> 1 t h e e m i s s i o n i s p r e d o m i n a n t l y due t o s p o n t a n e o u s t r a n s i t i o n s and i s t h u s i n c o h e r e n t ” )

16 end

147

Scilab code Exa 23.31.5 To determine coherence time and the coherence length 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y 4 clc 5 clear 6 u =8/(10^14) // u=(V/ v )=t h e s h o r t term f r e q u e n c y 7 8 9 10 11 12 13 14 15 16

s t a b i l i t y o f a He−Ne g a s l a s e r // v=c / y where c= v e l o c i t y o f l i g h t i n vacuum and y= wavelength c =3*10^8 // c= v e l o c i t y o f l i g h t i n m/ s y =1153*10^( -9) // y=e m i t t e d w a v e l e n g t h i n m e t e r s V =( u * c ) / y t =1/ V // t=c o h e r e n c e t i m e disp ( ”ms” ,( t *(10^3) ) ,” The c o h e r e n c e t i m e i s =” ) format ( ” v ” ,9) Lc = c / V // Lc=c o h e r e n c e l e n g t h disp ( ”m” ,Lc , ” The c o h e r e n c e l e n g t h i s =” ) format ( ” v ” ,3)

Scilab code Exa 23.31.6 To find the line width and the coherence length 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y 4 clc 5 clear 6 // y0=vacuum w a v e l e n g t h f o r t h e f r e q u e n c y v 7 // c =(v ∗ y0 )

148

// The d e v i a t i o n i n t h e w a v e l e n g t h i s Y0=( c ∗V) / ( v ˆ 2 ) //Y0=(( y0 ˆ 2 ) ∗V) / c //V b e i n g s p r e a d i n f r e q u e n c y o v e r t h e l i n e w i d t h . //V=(1/ t c ) c =3*(10^8) // c= v e l o c i t y o f l i g h t i n m/ s tc =10^( -8) // t c=c o h e r e n c e t i m e i n s e c o n d s y0 =650*(10^( -9) ) // y0=vacuum w a v e l e n g t h i n m Y0 =( y0 ^2) /( c * tc ) format ( ” e ” ,9) disp ( ”nm” , Y0 /(10^ -9) ,” L i n e w i d t h i s =” ) //Y0 i s c o n v e r t e d i n t e r m s o f nm 18 Lc = c * tc // Lc=c o h e r e n c e l e n g t h 19 format ( ” v ” ,5) 20 disp ( ”m” ,Lc , ” The c o h e r e n c e l e n g t h Lc i s =” )

8 9 10 11 12 13 14 15 16 17

Scilab code Exa 23.31.7 To find the radius along with the power density of the image and the coherence length 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

// s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y clc clear o =5*10^ -5 // o=a n g u l a r s p r e a d i n r a d i a n s f =10 // f= f o c a l l e n g t h i n cm D = f * o //D=d i a m e t e r o f t h e image r =( D /2) // r=image r a d i u s format ( ” v ” ,15) disp ( ”cm” ,r , ” The image r a d i u s i s =” ) a = %pi *( r ^2) // a=c r o s s s e c t i o n a l a r e a o f t h e image i n cmˆ2 P =10*10^ -3 //P=power i n Watts PD = P / a //PD=power d e n s i t y format ( ” v ” ,10) disp ( ”W/cmˆ2 ” ,PD , ” Power d e n s i t y i s =” ) 149

17 y =6000*10^ -8 // y=w a v e l e n g t h i n cm 18 d = y / o // d=c o h e r e n t w i d t h 19 disp ( ”cm” ,d , ” The l a t e r a l c o h e r e n t w i d t h

i s =” )

Scilab code Exa 23.31.8 To find the amount of pumping energy required for transition from 3s to 2p // s c i l a b 5 . 4 . 1 // Windows 7 o p e r a t i n g s y s t e m // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y clc clear h =6.62*10^ -34 // h=Planck ’ s c o n s t a n t c =3*10^8 // c= v e l o c i t y o f l i g h t i n vacuum i n m/ s y =632.8*10^ -9 // y=e m i t t e d w a v e l e n g t h i n m E =( h * c ) / y //E=e m i t t e d p h o t o n e n e r g y i n J o u l e s e =15.2*10^ -19 // e=e n e r g y o f 2 p l e v e l i n J o u l e s P = E + e //P=Pumping e n e r g y r e q u i r e d f o r t r a n s i t i o n from 3 s t o 2p l e v e l i n a He−Ne l a s e r 12 disp ( ”eV” ,P /(1.6*10^ -19) ,” The d e s i r e d pumping e n e r g y i s =” ) 13 format ( ” v ” ,10) 1 2 3 4 5 6 7 8 9 10 11

Scilab code Exa 23.31.9 To calculate the probability of stimulated emission 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y 4 clc 5 clear 6 h =6.62*10^ -34 // h=Planck ’ s c o n s t a n t 7 v =2.4*10^15 // v=f r e q u e n c y o f e m i t t e d r a d i a t i o n i n Hz

150

8 c =3*10^8 // c= v e l o c i t y o f l i g h t i n vacuum i n m/ s 9 A21 =1/(1.66*10^ -8) // A21=mean s p o n t a n e o u s l i f e t i m e 10 B21 =(( c ^3) * A21 ) /(8* %pi * h *( v ^3) ) // B21= p r o b a b i l i t y o f 11

stimulated emission disp ( ”mˆ 3 / ( J . s ˆ 2 ) ” ,B21 , ” The d e s i r e d p r o b a b i l i t y i s = ”)

Scilab code Exa 23.31.10 To calculate the NA and the acceptance angle along with number of reflections per metre 1 // s c i l a b 5 . 4 . 1 2 // Windows 7 o p e r a t i n g s y s t e m 3 // c h a p t e r 23 L a s e r s , F i b r e O p t i c s , and H o l o g r a p h y 4 clc 5 clear 6 u1 =1.55 // u1= r e f r a c t i v e i n d e x o f t h e c o r e o f t h e 7 8 9 10 11 12 13 14 15 16 17 18

fibre u2 =1.50 // u2= r e f r a c t i v e i n d e x o f t h e c l a d d i n g oa = asind ( sqrt (( u1 ^2) -( u2 ^2) ) ) // oa=a c c e p t a n c e a n g l e format ( ” v ” ,5) disp ( ” d e g r e e ” ,oa , ” The a c c e p t a n c e a n g l e i s =” ) NA = sind ( oa ) //NA=n u m e r i c a l a p e r t u r e disp ( , NA , ”NA=” ) oc = asind ( u2 / u1 ) // o c= c r i t i c a l a n g l e disp ( ” d e g r e e ” ,oc , ” C r i t i c a l a n g l e=” ) d =50*10^ -6 // d=c o r e d i a m e t e r i n m e t e r s x = d * tand ( oc ) // x= a x i a l d i s t a n c e t r a v e r s e d by t h e r a y b e t w e e n two c o n s e c u t i v e r e f l e c t i o n s n =1/ x // n=number o f r e f l e c t i o n s p e r m e t r e disp ( ,n , ” The number o f r e f l e c t i o n s p e r m e t r e i s =” )

151

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